QUESTION 7 PSYCH STATISTICS 7. (10 pts) I flip a fair coin 17 times. Answer the
ID: 3328376 • Letter: Q
Question
QUESTION 7 PSYCH STATISTICS
7. (10 pts) I flip a fair coin 17 times. Answer the following questions times. A a. What is the probability of getting 9 heads? b. What is the probability of getting 2 heads? c. What is the probability of getting 1 tail? d. What is the probability of getting 14 or more heads? e. What is the probability of getting 17 tails? I toss an unfair coin 12 times. This coin is 65% likely to show up heads. probability of the following. a. 11 heads: b. 2 or more heads: c. 7 heads: d. 9 tails: e. 8 or less heads: (4 pts) a. If the average American sleeps 8 hours a night, with a stan deviation of 1 hour, and I plan on gathering a sample of 12 college compare to this population, find the following: 8.Explanation / Answer
7) p = 0.5
n = 17
P(X = x) = 17Cx * 0.5x * 0.517-x
a) P(X = 9) = 17C9 * 0.59 * 0.58 = 0.1855
b) P(X = 2) = 17C2 * 0.52 * 0.515 = 0.0010
c) P(1 tail) = P(16 heads) = P(X = 16) = 17C16 * 0.516 * 0.51 = 0.0001
d) P(X > 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17)
= 17C14 * 0.514 * 0.53 + 17C15 * 0.515 * 0.52 + 17C16 * 0.516 * 0.51 + 17C17 * 0.517 * 0.50
= 0.0064
e) P(17 tails) = P(0 heads) = P(X = 0) = 17C0 * 0.50 * 0.517 = 0
P(head) = 0.65
P(tails) = 1 - 0.65 = 0.35
n = 12
a) P(11 heads) = 12C11 * 0.6511 * 0.351 = 0.0368
b) P(2 or more heads) = P(X > 2)
= 1 - [P(X = 0) + P(X = 1)]
= 1 - [12C0 * 0.650 * 0.3512 + 12C1 * 0.651 * 0.3511]
= 1 - 0.0001
= 0.9999
c) P(7 heads) = 12C7 * 0.657 * 0.355 = 0.2039
d) P(9 tails) = 12C9 * 0.359 * 0.653 = 0.0048
e) P(8 or less heads) = P(X < 8)
= 1 - P(X > 8)
= 1 - [P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)]
= 1 - [12C9 * 0.659 * 0.353 + 12C10 * 0.6510 * 0.352 + 12C11 * 0.6511 * 0.351 + 12C12 * 0.6512 * 0.350]
= 1 - 0.0056
= 0.9944
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